CBSE Class 10 Science: 2025 board paper, set 31/4/1

1. Souring of milk, fermentation of grapes and rusting of iron all form new substances, so they are chemical changes.

2. Melting of glaciers only changes ice to water (a change of state). No new substance forms, so it is a physical change. Option (D).

   1 mark

Answer: (D) Melting of glaciers

1. The guard tube needs a drying agent that absorbs moisture but does not react with HCl gas.

2. Anhydrous calcium chloride is a neutral drying agent and does not react with HCl. CaO, Ca(OH)${}_2$ and CaCO${}_3$ are basic and would react with the acid. Option (A).

   1 mark

Answer: (A) Calcium chloride

1. Ductility is the ability of a material to be drawn into thin wires. Malleability is beating into sheets. Option (B).

   1 mark

Answer: (B) ductility

1. Propyne is $C{H}_3-C\equiv CH$. It has three carbons joined by exactly one C–C single bond and one C≡C triple bond.

2. Butyne has four carbons, so it has two C–C bonds. Benzene and cyclohexane have no triple bond. Option (D).

   1 mark

Answer: (D) Propyne

1. Proteins (amino acids) contain nitrogen. Plants absorb it from the soil as nitrates or nitrites. Option (B).

   1 mark

Answer: (B) Nitrogen

1. Lymph contains some plasma, proteins and blood cells (mainly lymphocytes) but no RBCs. So B is true and C is false.

2. Lymph vessels drain into larger veins, not arteries. So D is true and A is false. The answer is B and D, option (B).

   1 mark

Answer: (B) B and D

1. Rose and banana have lost the capacity to produce seeds. They are propagated vegetatively, for example by cuttings and suckers. Option (C).

   1 mark

Answer: (C) seeds

1. The anther of the stamen produces pollen grains, which contain the male gametes. Option (A).

   1 mark

Answer: (A) anther

1. $TT\times tt$ gives an F1 of all $Tt$, so 100% of plants are tall.

2. Selfing F1 ($Tt\times Tt$) gives $TT:Tt:tt=1:2:1$, so 3/4 = 75% of plants are tall. Option (C).

   1 mark

Answer: (C) 100% ; 75%

1. A real image with $m=-1$ (same size, inverted) is formed by a convex lens when the object is at $2F$.

2. So $u=2f=2\times 20=40$ cm. Option (C).

   1 mark

Answer: (C) 40 cm

1. In a glass slab, the emergent ray is parallel to the incident ray but shifted sideways. This shift is called lateral displacement. Option (C).

   1 mark

Answer: (C) glass slab

1. Each of the three parts has resistance $R/3$.

2. In parallel, $R^{\prime }=\frac{R/3}{3}=R/9$, so $R/R^{\prime }=9$. Option (D).

   1 mark

Answer: (D) 9

1. Use $R=\frac{V^2}{P}=\frac{220^2}{11}=\frac{48400}{11}=4400\Omega$. Option (A).

   1 mark

Answer: (A) 4400 Ω

1. In series, the supply voltage is shared equally, so each bulb must get 4 V.

2. Number of bulbs $n=240/4=60$. Option (C).

   1 mark

Answer: (C) 60

1. Only about 10% of energy passes from one trophic level to the next, so the fewer the levels, the more energy is available at the top.

2. Plants → Man has only two trophic levels, so it loses the least energy. Option (C).

   1 mark

Answer: (C) Plants → Man

1. DDT is non-biodegradable. It is not broken down or excreted, so its concentration increases at each trophic level. Option (C).

   1 mark

Answer: (C) DDT

1. A is false: in large animals, diffusion alone cannot deliver oxygen to all body parts easily.

2. R is true: respiratory pigments such as haemoglobin take up oxygen and carry it to the tissues. Option (D).

   1 mark

Answer: (D) A false, R true

1. A is false: an acid is diluted by adding the acid slowly to water, never water to acid, because the process is highly exothermic.

2. R is true: concentrated nitric acid dissolves readily in water. Option (D).

   1 mark

Answer: (D) A false, R true

1. Both statements are true: in some reptiles, such as turtles, the incubation temperature of the eggs decides the sex of the offspring.

2. This happens because sex is not genetically determined in these animals, so R correctly explains A. Option (A).

   1 mark

Answer: (A)

1. A is false: when the ciliary muscles contract, the lens becomes thicker; it becomes thin when they relax.

2. R is true: by changing the curvature of the lens, the ciliary muscles control its power. Option (D).

   1 mark

Answer: (D) A false, R true

1. Oxidation is a reaction in which a substance gains oxygen or loses hydrogen (in terms of electrons, it loses electrons).

   1 mark

2. In $\mathrm{CuO}+{\mathrm{H}}_2\to \mathrm{Cu}+{\mathrm{H}}_2\mathrm{O}$, hydrogen gains oxygen to form water. So the substance oxidised is **hydrogen (${\mathrm{H}}_2$)**. CuO loses oxygen and is reduced.

   1 mark

Answer: Oxidation is gain of oxygen or loss of hydrogen; the substance oxidised is hydrogen (${\mathrm{H}}_2$).

1. Electronic configurations: Mg (12) = 2, 8, 2; Cl (17) = 2, 8, 7.

   Mg gives one valence electron to each of two Cl atoms, so that every ion gets a complete octet:

   $\mathrm{M}{\mathrm{g}}^{\times \times }+2\cdot \ddot{\underset{..}{\mathrm{Cl}}}:\to \mathrm{M}{\mathrm{g}}^{2+}+2[{}^{\times }\ddot{\underset{..}{\mathrm{Cl}}}:{]}^{-}$

   That is, $\mathrm{Mg}\to \mathrm{M}{\mathrm{g}}^{2+}+2{\mathrm{e}}^{-}$ and $2Cl+2{\mathrm{e}}^{-}\to 2C{\mathrm{l}}^{-}$, which together give $\mathrm{MgC}{\mathrm{l}}_2$ (that is, $\mathrm{M}{\mathrm{g}}^{2+}(\mathrm{C}{\mathrm{l}}^{-}{)}_2$).

   1 mark

2. Cation: magnesium ion, $\mathrm{M}{\mathrm{g}}^{2+}$.

   ½ mark

3. Anion: chloride ion, $\mathrm{C}{\mathrm{l}}^{-}$.

   ½ mark

Answer: $\mathrm{Mg}\to \mathrm{M}{\mathrm{g}}^{2+}+2{\mathrm{e}}^{-}$, $2Cl+2{\mathrm{e}}^{-}\to 2C{\mathrm{l}}^{-}$, giving $\mathrm{MgC}{\mathrm{l}}_2$; cation $\mathrm{M}{\mathrm{g}}^{2+}$, anion $\mathrm{C}{\mathrm{l}}^{-}$.

1. Zinc is in the middle of the activity series. Its sulphide or carbonate ore is first changed into the oxide, and the oxide is then reduced with carbon.

2. Roasting (for the sulphide ore, zinc blende): the ore is heated strongly in excess air.

   $2ZnS(\mathrm{s})+3{\mathrm{O}}_2(\mathrm{g})\stackrel{\mathrm{Heat}}{\to }2ZnO(\mathrm{s})+2S{\mathrm{O}}_2(\mathrm{g})$

   1 mark

3. Reduction: the zinc oxide is heated with carbon (coke).

   $\mathrm{ZnO}(\mathrm{s})+\mathrm{C}(\mathrm{s})\stackrel{\mathrm{Heat}}{\to }\mathrm{Zn}(\mathrm{s})+\mathrm{CO}(\mathrm{g})$

   1 mark

4. Alternatively, for the carbonate ore the first step is calcination (strong heating in limited air): $\mathrm{ZnC}{\mathrm{O}}_3(\mathrm{s})\stackrel{\mathrm{Heat}}{\to }\mathrm{ZnO}(\mathrm{s})+\mathrm{C}{\mathrm{O}}_2(\mathrm{g})$. This is followed by the same reduction with carbon.

Answer: Roasting: $2ZnS+3{\mathrm{O}}_2\to 2ZnO+2S{\mathrm{O}}_2$ (or calcination: $\mathrm{ZnC}{\mathrm{O}}_3\to \mathrm{ZnO}+\mathrm{C}{\mathrm{O}}_2$), then reduction: $\mathrm{ZnO}+\mathrm{C}\to \mathrm{Zn}+\mathrm{CO}$.

1. Oxygen produced in photosynthesis is given out as a waste through the stomata.

   ½ mark

2. Excess water is removed by transpiration.

   ½ mark

3. Many wastes collect in old leaves, which are then shed.

   ½ mark

4. Some wastes are stored as resins and gums, especially in old xylem. Other possible points: wastes are stored in cell vacuoles, or excreted into the soil around the roots.

   ½ mark

Answer: Oxygen through stomata, excess water by transpiration, shedding of leaves, and storage as resins and gums in old xylem (also storage in vacuoles or excretion into the soil).

1. Flow chart

   Father (XY) → sperms: 50% carry X, 50% carry Y

   Mother (XX) → all ova carry X

   Sperm (X) + Ovum (X) → zygote XX → girl

   Sperm (Y) + Ovum (X) → zygote XY → boy

   1 mark

2. The mother can only give an X chromosome, but the father can give either X or Y. So the sex of the child depends on which type of sperm fertilises the ovum. Therefore the father is responsible for the sex of the child.

   1 mark

Answer: X-sperm + X-ovum gives XX (girl); Y-sperm + X-ovum gives XY (boy). Only the father can contribute Y, so the father decides the sex of the child.

1. Ray diagram (described): Draw an equilateral triangle PQR as the prism. A ray EF from the left strikes face PQ at F and bends towards the normal on entering glass. The refracted ray FG travels inside the prism and meets face PR at G. There it bends away from the normal and emerges as ray GH.

   1 mark

2. Labels: incident ray, refracted ray, emergent ray, the normals at F and G, angle of incidence $\angle i$, angle of refraction $\angle r$, angle of emergence $\angle e$, and the angle of the prism $\angle A$ at P.

   ½ mark

3. Extend the incident ray EF forward and the emergent ray GH backward as dotted lines until they meet. The angle between them, $\angle D$, is marked and named the angle of deviation.

   ½ mark

Answer: Ray bends towards the base at both faces; the angle between the extended incident ray and the emergent ray is the angle of deviation ($\angle D$).

1. Persons with presbyopia often have both myopia and hypermetropia. They need bifocal lenses.

   ½ mark

2. Structure: the upper portion of a bifocal lens is a concave lens and the lower portion is a convex lens.

   1 mark

3. Reason: the upper concave part helps in seeing distant objects, and the lower convex part helps in seeing nearby objects, for example while reading. (If only near vision is affected, a convex lens is used.)

   ½ mark

Answer: Bifocal lenses: concave upper part for distant vision and convex lower part for near vision.

1. Magnetic field lines are imaginary lines drawn around a magnet to represent its magnetic field. The tangent at any point on a line gives the direction of the field at that point.

   1 mark

2. Property 1: No two magnetic field lines ever cross each other.

   ½ mark

3. Property 2: Outside the magnet, field lines emerge from the north pole and merge at the south pole. They are closed curves, and inside the magnet they run from the south pole to the north pole.

   ½ mark

Answer: Magnetic field lines are imaginary lines representing the magnetic field around a magnet. They never cross, and they run from N to S outside the magnet as closed curves.

1. We balance a chemical equation so that the number of atoms of each element is the same on both sides of the reaction.

   ½ mark

2. The law is the Law of Conservation of Mass.

   ½ mark

3. Statement: Mass can neither be created nor destroyed in a chemical reaction. The total mass of the products equals the total mass of the reactants.

   1 mark

4. Balance the $\mathrm{P}{\mathrm{O}}_4$ groups and Zn first, then hydrogen:

   $3Zn+2{\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_4\to \mathrm{Z}{\mathrm{n}}_3(\mathrm{P}{\mathrm{O}}_4{)}_2+3{\mathrm{H}}_2$

   Check: Zn 3 = 3, P 2 = 2, O 8 = 8, H 6 = 6.

   1 mark

Answer: To make the atoms of each element equal on both sides; Law of Conservation of Mass; $3Zn+2{\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_4\to \mathrm{Z}{\mathrm{n}}_3(\mathrm{P}{\mathrm{O}}_4{)}_2+3{\mathrm{H}}_2$.

1. A precipitation reaction is a reaction in which an insoluble substance (precipitate) is formed when solutions of two compounds are mixed.

   1 mark

2. Example: When sodium sulphate solution is added to barium chloride solution, a white precipitate of barium sulphate is formed.

   1 mark

3. $\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4(\mathrm{aq})+\mathrm{BaC}{\mathrm{l}}_2(\mathrm{aq})\to \mathrm{BaS}{\mathrm{O}}_4(\mathrm{s})\downarrow +2NaCl(\mathrm{aq})$

   1 mark

Answer: A reaction that forms an insoluble precipitate; e.g. $\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4(\mathrm{aq})+\mathrm{BaC}{\mathrm{l}}_2(\mathrm{aq})\to \mathrm{BaS}{\mathrm{O}}_4(\mathrm{s})+2NaCl(\mathrm{aq})$ (white ppt).

1. Set-up: Take an aluminium or copper wire and clamp it to a stand. Fix a pin to its free end using a little wax.

   1 mark

2. Procedure and observation: Heat the wire with a spirit lamp or burner near the clamped end. After some time the wax melts and the pin falls off, but the wire itself does not melt.

   1 mark

3. Conclusion: Heat travelled along the wire to the far end and melted the wax, so metals are good conductors of heat. The wire did not melt even in the flame, so metals have high melting points.

   1 mark

Answer: Heat a clamped metal wire with a pin fixed by wax at its far end. The pin falls off but the wire does not melt, so metals conduct heat well and have high melting points.

1. (i) Saliva: contains the enzyme salivary amylase.

   ½ mark

2. (i) Function: salivary amylase breaks down starch into simple sugar.

   ½ mark

3. (ii) Bile juice: contains bile salts.

   ½ mark

4. (ii) Function: bile salts emulsify fats into small globules, and bile makes the acidic food from the stomach alkaline so that pancreatic enzymes can act.

   ½ mark

5. (iii) Pancreatic juice: contains the enzymes trypsin and lipase.

   ½ mark

6. (iii) Function: trypsin digests proteins, and lipase breaks down emulsified fats.

   ½ mark

Answer: (i) Salivary amylase: starch to sugar; (ii) Bile salts: emulsify fats and make food alkaline; (iii) Trypsin: digests proteins / Lipase: breaks down emulsified fats.

1. Limitation 1: Electrical impulses reach only those cells that are connected by nervous tissue, not every cell of the body.

   1 mark

2. Limitation 2: Once a cell generates and transmits an impulse, it takes some time to reset its mechanism before it can send a new impulse. So cells cannot generate and transmit impulses continuously.

   1 mark

3. Chemical communication is better: hormones are released into the blood, so they can potentially reach all cells of the body. They act steadily and persistently to bring about the desired changes, such as growth.

   1 mark

Answer: Impulses reach only nerve-connected cells, and cells need time to reset before the next impulse. Chemical signals (hormones) reach all cells steadily and persistently.

1. A concave mirror forms a virtual, erect and magnified image only when the object is placed between the pole P and the focus F, that is, at less than 18 cm from the mirror.

   1 mark

2. Mirror formula: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$, with $f=-18$ cm.

   ½ mark

3. Magnification $m=-\frac{v}{u}=+2$, so $v=-2u$.

   1 mark

4. Substituting: $\frac{1}{-2u}+\frac{1}{u}=\frac{1}{-18}$

   $\frac{1}{2u}=-\frac{1}{18}$

   $u=-9$ cm, and then $v=+18$ cm (a virtual image behind the mirror).

   The object is 9 cm in front of the mirror, which is less than $f=18$ cm. It lies between P and F, which justifies the answer.

   ½ mark

Answer: Object between P and F (less than 18 cm from the mirror); $u=-9$ cm.

1. (i) A ($10^{17}$ Ω m, extremely high resistivity) is an insulator.

   ½ mark

2. (i) B ($44\times 10^{-6}$ Ω m, moderately high) is an alloy.

   ½ mark

3. (i) C ($1.62\times 10^{-8}$ Ω m, very low) is a conductor.

   ½ mark

4. (ii) A, insulator: plastic or rubber, used for the handle of an electric iron or the feet of an electric stove.

   ½ mark

5. (ii) B, alloy: nichrome, used as the heating element of an electric iron or stove.

   ½ mark

6. (ii) C, conductor: copper, used for the connecting wires and cord of the appliance.

   ½ mark

Answer: (i) A insulator, B alloy, C conductor; (ii) plastic for the iron handle, nichrome for the heating element, copper for the connecting wires.

1. Decomposers are microorganisms that break down the complex organic substances in dead plants and animals into simple inorganic substances.

   1 mark

2. Examples: bacteria and fungi.

   1 mark

3. The simple substances they release go into the soil and are used again by plants. This recycles nutrients, cleans away dead matter and keeps the ecosystem in balance.

   1 mark

Answer: Decomposers (e.g. bacteria and fungi) break down dead organic matter into simple substances that return to the soil for reuse by plants, maintaining balance in the ecosystem.

1. 'C' burns to give 2 mol $\mathrm{C}{\mathrm{O}}_2$ and 3 mol ${\mathrm{H}}_2\mathrm{O}$, so it has 2 C and 6 H atoms. 'C' is **ethane, ${\mathrm{C}}_2{\mathrm{H}}_6$**.

   Structure: ${\mathrm{H}}_3\mathrm{C}-\mathrm{C}{\mathrm{H}}_3$, i.e. each carbon bonded to three H atoms and to the other carbon by a single bond.

   1 mark

2. 'B' adds one mole of ${\mathrm{H}}_2$ to give ethane, so 'B' is **ethene, ${\mathrm{C}}_2{\mathrm{H}}_4$**.

   Structure: ${\mathrm{H}}_2\mathrm{C}=\mathrm{C}{\mathrm{H}}_2$ (two carbons joined by a double bond, each carrying two H atoms).

   1 mark

3. 'B' is formed from 'A' by dehydration, so 'A' is **ethanol, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$**.

   Structure: ${\mathrm{H}}_3\mathrm{C}-\mathrm{C}{\mathrm{H}}_2-\mathrm{O}-\mathrm{H}$.

   1 mark

4. A → B: $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\mathrm{Hot}conc.{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}$

   ½ mark

5. B → C: $\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\stackrel{\mathrm{Ni}}{\to }\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}_3$

   ½ mark

6. Combustion of C: ${\mathrm{C}}_2{\mathrm{H}}_6+\frac{7}{2}{\mathrm{O}}_2\to 2C{\mathrm{O}}_2+3{\mathrm{H}}_2\mathrm{O}$ (or $2{\mathrm{C}}_2{\mathrm{H}}_6+7{\mathrm{O}}_2\to 4C{\mathrm{O}}_2+6{\mathrm{H}}_2\mathrm{O}$)

   ½ mark

7. **Role of conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$:** it acts as a dehydrating agent and removes a water molecule from ethanol.

   ½ mark

Answer: A = ethanol ($\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$), B = ethene ($\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2$), C = ethane (${\mathrm{C}}_2{\mathrm{H}}_6$); conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ acts as a dehydrating agent.

1. (i) 'A' is **ethanoic acid (acetic acid), $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$**. A dilute solution of it (vinegar) is used as a pickle preservative.

   ½ mark

2. (i) Structure: a $\mathrm{C}{\mathrm{H}}_3$ group is attached to a carbon that is double-bonded to O and single-bonded to an OH group, i.e. ${\mathrm{H}}_3\mathrm{C}-\mathrm{C}(=\mathrm{O})-\mathrm{O}-\mathrm{H}$.

   ½ mark

3. (ii) Esterification gives the sweet-smelling ester 'B', ethyl ethanoate:

   $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\stackrel{\mathrm{Acid}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$

   ½ mark

4. (ii) Role of acid: concentrated sulphuric acid acts as a catalyst.

   ½ mark

5. (iii) Heat the ester 'B' with dilute NaOH (saponification):

   $\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+\mathrm{NaOH}\to \mathrm{C}{\mathrm{H}}_3\mathrm{COONa}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$

   Acidifying the sodium ethanoate gives ethanoic acid 'A' back. Alternatively, hydrolyse the ester with water in the presence of an acid.

   1 mark

6. (iv) Oxidise ethanol by warming it with alkaline potassium permanganate or acidified potassium dichromate:

   $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{Alk.\mathrm{KMn}{\mathrm{O}}_4+\mathrm{Heat}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$

   1 mark

7. (v) **Carbon dioxide ($\mathrm{C}{\mathrm{O}}_2$)** is produced:

   $2C{\mathrm{H}}_3\mathrm{COOH}+\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3\to 2C{\mathrm{H}}_3\mathrm{COONa}+{\mathrm{H}}_2\mathrm{O}+\mathrm{C}{\mathrm{O}}_2$

   1 mark

Answer: (i) Ethanoic acid, $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$; (ii) $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\to \mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$, acid acts as catalyst; (iii) hydrolysis/saponification of the ester with NaOH, then acidification; (iv) oxidation with alkaline $\mathrm{KMn}{\mathrm{O}}_4$ or acidified ${\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7$; (v) $\mathrm{C}{\mathrm{O}}_2$.

1. (i) Regeneration is the ability of a fully differentiated organism to give rise to new individuals from its body parts. If the organism is cut into pieces, each piece grows into a complete organism.

   1 mark

2. (i) An organism that shows regeneration: Planaria (or Hydra).

   ½ mark

3. (i) An organism that does not show regeneration: Spirogyra.

   ½ mark

4. (i) Reason: regeneration needs specialised cells that can proliferate and form different cell types and tissues. Spirogyra is a simple filament without such specialised cells, so it cannot regenerate.

   1 mark

5. (ii) The filamentous structures are Spirogyra (a green alga).

   ½ mark

6. (ii) They reproduce by fragmentation.

   ½ mark

7. (ii) Process: on maturing, the Spirogyra filament simply breaks up into smaller pieces (fragments). Each fragment grows into a new individual filament.

   1 mark

Answer: (i) Regeneration means forming new organisms from body parts; Planaria regenerates, Spirogyra does not because it lacks specialised cells. (ii) Spirogyra reproduces by fragmentation: the mature filament breaks into pieces that each grow into a new individual.

1. (i)(a) Carries sperms: vas deferens.

   ½ mark

2. (i)(b) Produces male gametes: testes.

   ½ mark

3. (i)(c) Secretion makes the transport of sperms easier: prostate gland and seminal vesicles.

   ½ mark

4. (i)(d) Provides a suitable (lower) temperature for sperm formation: scrotum.

   ½ mark

5. (ii) Sperms are tiny cells that consist mainly of genetic material. Each sperm has a long tail that helps it move towards the female germ cell.

   1 mark

6. (iii) Surgical contraceptive methods block the gamete-carrying tube: the vas deferens in males (vasectomy) or the fallopian tube in females (tubectomy). This prevents sperm and egg from meeting, so fertilisation does not occur.

   1 mark

7. (iii) Side effect: if the surgery is not performed properly, it can cause infections and other problems.

   1 mark

Answer: (i)(a) Vas deferens (b) Testes (c) Prostate gland/seminal vesicles (d) Scrotum; (ii) small in size, contain genetic material, have a tail for movement; (iii) blocking the vas deferens or fallopian tube by surgery; it can cause infection if not done properly.

1. (i) Pattern: Around each conductor the field lines are concentric circles lying in the plane of the cardboard. They are closer together near each wire and spread out farther away. Between the two wires the lines crowd together.

   1 mark

2. (i) Directions (right-hand thumb rule): The left conductor carries current into the cardboard (⊗), so its circles run clockwise. The right conductor carries current out of the cardboard (⊙), so its circles run anticlockwise. At the midpoint O, both fields point the same way: perpendicular to the line joining the conductors, towards the bottom of the figure. The two fields therefore add, and the resultant field at O is strong and points downward in the plane of the cardboard. Show this with an arrow at O.

   1 mark

3. (ii)(a) Red insulation – live wire. Black insulation – neutral wire. Green insulation – earth wire.

   1½ marks

4. (ii)(b) The potential difference between the live (red) and neutral (black) wires is 220 V.

   ½ mark

5. (ii)(c) The earth wire is a safety measure. It connects the metallic body of the appliance to a metal plate buried deep in the earth. If current leaks to the body, the body stays at earth's potential (zero) and the leaked current flows to the earth. So the user does not get a severe electric shock.

   1 mark

Answer: (i) Concentric circles: clockwise around the left (inward) wire, anticlockwise around the right (outward) wire; the fields add at O, pointing perpendicular to the line joining the wires. (ii)(a) Red – live, black – neutral, green – earth; (b) 220 V; (c) Earth wire keeps the metal body at zero potential and drains leaked current, preventing a severe shock.

1. (i)(a) Place rod AB between the poles of the horseshoe magnet and close key K so that current flows through AB. The rod is seen to be displaced (it moves). This shows that a force acts on the current-carrying conductor in the magnetic field.

   1 mark

2. (i)(b) Way 1: Reverse the direction of the current in AB by reversing the battery connections. The rod is now displaced in the opposite direction.

   1 mark

3. (i)(b) Way 2: Reverse the direction of the magnetic field by interchanging the poles of the magnet (S on top, N below). The rod again moves the opposite way.

   1 mark

4. (ii) The force is highest when the direction of the current is at right angles (perpendicular) to the direction of the magnetic field.

   1 mark

5. (iii) Fleming's left-hand rule: Stretch the thumb, forefinger and middle finger of the left hand so that they are mutually perpendicular. If the forefinger points in the direction of the magnetic field and the middle finger in the direction of the current, then the thumb points in the direction of the force (motion) on the conductor.

   1 mark

Answer: (i)(a) AB gets displaced when current flows; (b) reverse the current, or reverse the magnetic field. (ii) When the current is perpendicular to the magnetic field. (iii) Forefinger – field, middle finger – current, thumb – force, with all three mutually perpendicular on the left hand.

1. (i) Acid: hydrochloric acid (HCl). Base: sodium hydroxide (NaOH). $\text{HCl}+\text{NaOH}\to \text{NaCl}+{\text{H}}_2\text{O}$

   1 mark

2. (ii) Sodium chloride is neutral (pH 7), because it is a salt of a strong acid (HCl) and a strong base (NaOH).

   1 mark

3. (iii)(A) When electricity is passed through brine, it decomposes (chlor-alkali process): $2\text{NaCl}(aq)+2{\text{H}}_2\text{O}(l)\to 2\text{NaOH}(aq)+{\text{Cl}}_2(g)+{\text{H}}_2(g)$

   ½ mark

4. (iii)(A) Products and where they form: chlorine gas at the anode, hydrogen gas at the cathode, and sodium hydroxide solution near the cathode.

   1½ marks

5. (iii)(B) Step 1: Brine is treated with ammonia and carbon dioxide, giving baking soda: $\text{NaCl}+{\text{H}}_2\text{O}+{\text{CO}}_2+{\text{NH}}_3\to {\text{NH}}_4\text{Cl}+{\text{NaHCO}}_3$

   ½ mark

6. (iii)(B) Step 2: Sodium hydrogencarbonate is heated to give sodium carbonate: $2{\text{NaHCO}}_3\stackrel{\text{Heat}}{\to }{\text{Na}}_2{\text{CO}}_3+{\text{H}}_2\text{O}+{\text{CO}}_2$

   ½ mark

7. (iii)(B) Step 3: Sodium carbonate is recrystallised from water to give washing soda.

   ½ mark

8. (iii)(B) ${\text{Na}}_2{\text{CO}}_3+10{\text{H}}_2\text{O}\to {\text{Na}}_2{\text{CO}}_3\cdot 10{\text{H}}_2\text{O}$ (washing soda)

   ½ mark

Answer: (i) HCl and NaOH; (ii) Neutral – salt of a strong acid and a strong base; (iii)(A) Brine decomposes: Cl₂ at the anode, H₂ at the cathode, NaOH near the cathode; (iii)(B) NaCl + H₂O + CO₂ + NH₃ → NH₄Cl + NaHCO₃; 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂ on heating; Na₂CO₃ + 10H₂O → Na₂CO₃·10H₂O.

1. (i) Reflex action. It is a sudden, automatic and involuntary response to a stimulus in the environment, done without conscious thinking and controlled by the spinal cord.

   1 mark

2. (ii)(a) Motor neuron: carries the message (impulse) from the spinal cord (CNS) to the effector organ, i.e. the muscle, which then moves the hand.

   ½ mark

3. (ii)(b) Relay neuron: lies in the spinal cord and connects the sensory neuron to the motor neuron.

   ½ mark

4. (iii)(A) Central nervous system (CNS): its components are the brain and the spinal cord.

   1 mark

5. (iii)(A) Peripheral nervous system (PNS): its components are the cranial nerves (arising from the brain) and the spinal nerves (arising from the spinal cord).

   1 mark

6. (iii)(B)(a) Thinking – fore-brain (cerebrum).

   ½ mark

7. (iii)(B)(b) Picking up a pencil (precision of voluntary action) – cerebellum (hind-brain).

   ½ mark

8. (iii)(B)(c) Controlling blood pressure – medulla (hind-brain).

   ½ mark

9. (iii)(B)(d) Controlling hunger – fore-brain (it has a separate area for hunger).

   ½ mark

Answer: (i) Reflex action – a sudden, involuntary response to a stimulus; (ii)(a) Motor neuron carries the impulse from the spinal cord to the muscle; (b) Relay neuron links the sensory neuron to the motor neuron; (iii)(A) CNS – brain and spinal cord; PNS – cranial and spinal nerves; (iii)(B) (a) fore-brain/cerebrum, (b) cerebellum, (c) medulla, (d) fore-brain.

1. (i) A rainbow in the sky after a shower is a natural spectrum of sunlight.

   1 mark

2. (ii) The white light undergoes dispersion. Its component colours bend through different angles in the prism (violet the most, red the least), so it splits into the band VIBGYOR.

   1 mark

3. (iii)(A) Condition 1: tiny water droplets must be present in the atmosphere, for example just after rain.

   1 mark

4. (iii)(A) Condition 2: the Sun must be behind the observer, with the observer facing the droplets.

   1 mark

5. (iii)(B) Diagram: draw a circle for a raindrop and a ray of sunlight striking its upper-left surface. Inside the drop, show the light splitting into red and violet rays that travel to the back surface. There they reflect, then emerge from the lower-left surface heading downward, with red and violet separated, towards the observer.

   ½ mark

6. (iii)(B) Mark (a) at the point where sunlight enters the drop: refraction and dispersion occur here.

   ½ mark

7. (iii)(B) Mark (b) at the back surface of the drop: total internal reflection occurs here.

   ½ mark

8. (iii)(B) Mark (c) at the point where light leaves the drop: the final refraction occurs here, and the colours reach the eye.

   ½ mark

Answer: (i) Rainbow; (ii) Dispersion of white light into its seven colours; (iii)(A) Water droplets in the air, and the Sun behind the observer; (iii)(B) Raindrop diagram: (a) entry surface – dispersion, (b) back surface – internal reflection, (c) exit surface – final refraction.